| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Standard +0.8 This is a Further Maths question requiring product rule with arccos (part i) and chain rule with implicit differentiation involving hyperbolic functions (part ii). Part (i) is straightforward application, but part (ii) requires recognizing the connection between exponential and hyperbolic functions, computing a second derivative, and algebraic manipulation to reach the required form—this demands more sophistication than standard A-level calculus. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08g Derivatives: inverse trig and hyperbolic functions |
\begin{enumerate}[label=(\roman*)]
\item $f(x) = x \arccos x \quad -1 \leq x \leq 1$
Find the exact value of $f'(0.5)$.
[3]
\item $g(x) = \arctan(e^{2x})$
Show that
$$g''(x) = k \operatorname{sech}(2x) \tanh(2x)$$
where $k$ is a constant to be found.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q4 [8]}}